Properties of Turing Machines and NP-Completeness
Seminar 18:36:28 ChanServ changed the topic of #mathematics to: SEMINAR IN PROGRESS. If you want to ask a question, say ! and wait to be called 18:00:25 ~vixey: oh did I miss it 18:00:41 mark-t: nope, should be starting nowish 18:00:50 ~vixey: oh that's perfect! 18:02:58 mark-t: well, I'm starting now, with or without a topic change 18:04:02 mark-t: so, last time I gave a very brief introduction to turing machines and gave a somewhat precise statement of the P vs. NP problem 18:04:17 mark-t: but I don't feel like I really got across why anybody cares about this problem 18:05:04 mark-t: it might seem like turing machines are just some toy to be played with, and whether or not you can design fast ones to recognize certain languages may seem a rather trivial thing to worry about 18:05:46 mark-t: so, I'd like to spend a few minutes discussing how powerful turing machines are and why they're relevant to anything 18:06:08 mark-t: let's go back to my previous example: 18:06:44 mark-t: as you'll recall, this turing machine recognizes strings that have the same number of 'a's as 'b's 18:07:06 mark-t: it works in four steps: 18:07:15 mark-t: 1) find the first 'a' or 'b' in the string 18:07:31 mark-t: 2) if you found an 'a' in the string, cross it off, and find the next 'b' 18:07:50 mark-t: 3) if you found a 'b' in part 1), cross it off, and find the next 'a' 18:08:23 mark-t: 4) if you found a 'b' in 2) or an 'a' in 3), cross it off and rewind to the beginning of the string; repeat 18:09:19 mark-t: now, one thing you'll notice is that each of these steps is tied to a state in the machine 18:10:15 mark-t: so, while DFAs use states to record everything it knows about the string so far, turing machines use states to keep track of what step they're in in an algorithm, and they use the tape for memory 18:10:33 mark-t: in this way, turing machines are much more like CPUs than DFAs 18:11:35 mark-t: and that's how turing machines tend to be designed, as well; you think of steps to do a process, and you make a state for each step, instead of thinking about which strings are fundamentally different to the language and making them go to different states 18:12:30 mark-t: so, this is all summed up pretty well by the church-turing thesis, which states that any language that can be recognized by an effective algorithm can be recognized by a turing machine 18:13:04 mark-t: you can get more precise about that, but you should basically just have the idea that turing machines can compute anything you or I or your computer could compute 18:14:13 mark-t: and therefore, it should be no surprise that turing machines have equivalent power to non-deterministic turing machines; since we can trace things through and figure out if a non-deterministic turing machine accepts a given string, so can an ordinary turing machine 18:14:35 mark-t: ok, so that brings us to the question of P vs. NP 18:15:19 mark-t: one thing that strikes me about this problem is: ok, so you give the turing machine a string of length n; what on earth is it doing for an exponential amount of time? why is there anything that takes that long? 18:16:12 mark-t: well, it turns out that you can define languages with arbitrarily large time complexity; I'd like to show you how to create one 18:16:52 mark-t: for this example, I'll make one whose time complexity is at least O(2^n), but it will be quite obvious how to get the general case 18:17:10 mark-t: so, our alphabet for this example will just be {a} 18:17:55 mark-t: it should be pretty clear that the number of turing machines is countable; we need to get an ordered list of these and name them M_0, M_1, M_2, etc. 18:19:03 mark-t: so, to build this language, we start by checking the empty string, then "a", then "aa", then a^3, and so on 18:19:31 mark-t: at each step, we may mark a machine as canceled, which means that it cannot accept the language we're building 18:20:31 mark-t: so, here's how we check a^n; find the first uncanceled machine M_j, only checking machines up to M_n, that halts in less than 2^n steps when reading the string a^n 18:20:55 mark-t: if there isn't such a machine, don't add a^n to the language 18:21:46 mark-t: if there is such a machine, mark it as canceled; if it accepted the string, don't add a^n to the language; if it rejected the string, add a^n to the language 18:22:02 mark-t: and that's it 18:22:16 ~vixey: ? 18:22:19 mark-t: yes? 18:22:36 ~vixey: Do you have a picture of that machine? 18:22:51 mark-t: no; it would be quite a mess to construct, but this is all possible 18:23:39 mark-t: we can clearly associate a string to any turing machine; in order to make the list M_0, M_1, etc., we just have to generate them in order 18:24:01 mark-t: which could be something as inefficient as just generating every string in order and seeing if it's a valid turing machine 18:24:24 mark-t: we can clearly run any turing machine for 2^n steps and see if it accepts a given string 18:24:52 mark-t: and since we're only checking finitely many turing machines for each n, all of this will eventually finish 18:25:47 mark-t: I'll be avoiding the design of actual turing machines today, speaking more about what they can and can't do instead 18:26:10 mark-t: ok, so, let's show why this language has a time complexity at least O(2^n) 18:26:47 mark-t: first of all, we know there's some machine that can accept this language, since it can do all the steps I just described 18:27:19 mark-t: next, any machine that accepts this language must not be canceled in this process 18:27:42 mark-t: so, suppose we have a turing machine that accepts this language, and it's given by M_k 18:28:15 mark-t: by design, the only way that it could avoid being canceled is by taking 2^n steps or longer for every n >= k 18:28:46 mark-t: and therefore, for any machine that accepts this language, the time complexity is at least O(2^n) 18:29:35 mark-t: so, I hope it's abundantly clear how to make a language with arbitrarily large time complexity in this way, though it isn't necessarily clear what the machine is doing for all that time 18:29:55 mark-t: I can't necessarily help you with that 8) 18:30:29 mark-t: ok, back to P vs. NP 18:31:25 mark-t: one approach to this problem is looking at certain languages, which are called NP-complete 18:32:19 mark-t: a language in NP is called NP-complete if every language in NP can be reduced to it in polynomial time 18:32:50 mark-t: reduced means that you can transform the question about a string in one language to a question about a string in this language 18:33:46 mark-t: for example, if one language recognizes primes and another recognizes squares of primes, you reduce the former to the latter by squaring the input and checking if it's a square of a prime 18:34:40 mark-t: working in non-negative integers, of course 18:35:10 mark-t: first of all, it's not clear that an NP-complete language should exist; I'll get to that in a minute 18:35:59 mark-t: but if a language is NP-complete, then we know two things: if that language is in P, then P = NP; if that language is not in P, then P < NP 18:36:29 mark-t: of course, if we find any language in NP that isn't in P, then P < NP 18:37:06 mark-t: however, since that would imply that all of the NP-complete languages are not in P, the NP-complete languages are the obvious ones to look at 18:37:40 mark-t: you can think of an NP-complete language as being one of the hardest languages to recognize in NP 18:38:02 mark-t: now, back to the question of whether there even exists a language that is NP-complete 18:38:51 mark-t: here's the first example; the theorem that shows it's NP-complete is called Cook's Theorem 18:39:06 mark-t: the problem is the boolean satisfiability problem 18:39:30 mark-t: so, suppose you have a bunch of boolean variables; that is, each one takes a value of either true or false 18:40:33 mark-t: and you have a bunch of expressions in these variables, which consist of some of these variables and/or their complements "or"ed together 18:41:01 mark-t: for example, "~x1 v x2 v x3" (not-x1 or x2 or x3) would be such an expression 18:41:53 mark-t: the boolean satisfiability problem asks, given a set of these expressions, whether there some assignment to the variables that makes all of the expressions true 18:42:00 mark-t: is 18:43:22 mark-t: the proof of Cook's Theorem is rather extensive, so I won't go through it, but the main idea is that it takes the computation of a turing machine and expresses each step as a set of boolean expressions that all have to be true 18:44:11 mark-t: and then, if there's a solution to the set of boolean expressions, it means the turing machine will accept the language; if not, then it won't 18:44:52 mark-t: it shouldn't be too surprising that you can do this, as our computers express just about everything in terms of whether or not a given set of bits is true or false 18:45:27 mark-t: once you have one NP-complete language, finding others is much less work 18:45:55 mark-t: now, all we have to do is show that a given language is in NP and that some NP-complete language reduces to it in polynomial time 18:46:52 mark-t: since every language in NP will reduce to the NP-complete language in polynomial time, every language in NP could then be reduced to this language by reducing to the NP-complete language first 18:47:11 mark-t: oh, right, I forgot to mention why the boolean satisfiability problem (SAT) is in NP 18:48:13 mark-t: as you know, non-deterministic machines can do many computations in parallel; the implication is that they can generate all of the possible solutions and check each one in parallel 18:48:48 mark-t: and the time to generate a solution and check if it satisfies all of the boolean expressions is clearly polynomial in the length of the set of expressions 18:50:42 mark-t: this is a hallmark of NP-complete problems; there's generally some signature that can be computed to show when the answer is yes, and it can be checked in polynomial time 18:51:06 mark-t: the non-deterministic machine can just generate all of the possible signatures and check them in parallel 18:51:38 mark-t: the naive deterministic machine has to do them in series, which is why it's not clear whether or not they can do it in polynomial time 18:52:04 mark-t: ok, so here's another problem that is NP-complete 18:52:09 mark-t: this one is called 3SAT 18:52:33 mark-t: 3SAT is similar to SAT, except the expressions are required to have at most 3 variables each 18:53:12 mark-t: 3SAT is clearly in NP, since it can be checked by the same non-deterministic turing machine that checks SAT 18:53:30 mark-t: to show that 3SAT is NP-complete, we'll reduce SAT to it 18:54:21 mark-t: that is, we need to show that, given an expression from SAT with n variables, we can reduce it to a set of expressions with 3 variables that all have to be true 18:54:59 mark-t: so, let's say our expression was ~x1 v x2 v x3 v ~x4 v ... 18:55:59 mark-t: the idea is that we can replace the first two variables in the expression with a new variable, say y, and add an additional expression that makes sure y is only true when (~x1 v x2) was true 18:56:31 mark-t: by this process, we reduce the length of the original expression by 1 and add one new expression 18:56:52 mark-t: if the original expression had n variables, we'll have to do this n-3 times 18:57:31 mark-t: so, for the additional expression, we want to force y to be false unless ~x1 v x2 is true 18:57:47 mark-t: that is, ~y v ~x1 v x2 must be true 18:58:19 mark-t: 'cause if ~x1 v x2 is false, then we have to make y false in order to satisfy this expression 18:59:04 mark-t: and that's basically it; there's some analysis to show that this isn't going to create a set of expressions that's exponential in the length of the original set, but that's not hard to see 18:59:51 mark-t: you can't possibly be adding more than n expressions of fixed length when given a set of expressions containing n total variables 19:00:19 mark-t: therefore, we've reduced SAT to 3SAT; any SAT problem can be expressed as a 3SAT problem 19:01:36 mark-t: 3SAT is kind of a funny case; it's harder to reduce other problems to it, but it's easier to reduce 3SAT to other problems, so 3SAT is generally better than SAT for showing that another language is NP-complete 19:02:04 mark-t: there are lots of other NP-complete languages, but I don't have time to go into them 19:02:57 mark-t: some famous examples are the hamiltonian path problem, which is a special case of the traveling salesman problem, the subset sum problem, the vertex cover problem, and so on 19:03:26 mark-t: and, unfortunately, these problems are important in computer science, so we would really like to have fast algorithms to solve them 19:03:46 mark-t: that's why the P vs. NP problem is so important 19:03:56 mark-t: so, that's all I have planned for today 19:03:59 mark-t: thanks for reading 19:04:20 ~vixey: thank you! very interesting againa 19:04:22 tcoppi: :D thanks, nice talk 19:04:41 jamesstanley: so np-hard problems are problems that need to be brute forced? 19:05:03 mark-t: NP-hard problems are problems that any NP problem reduces to 19:05:07 jamesstanley: ok 19:05:12 mark-t: an NP-hard problem is NP-complete if it's in NP 19:05:23 ~vixey: I sort of want to see one of these turing machines 19:05:26 mark-t: it might be worse than NP, too 19:05:53 mark-t: the book I mentioned last time, "Languages and Machines" by Sudkamp does construct one for SAT 19:05:58 mark-t: I think 19:06:08 mark-t: maybe it was for the hamiltonian path problem 19:06:16 mark-t: it does one of these NP-complete problems 19:06:45 mark-t: but yes, generally the problem with np-hard problems is that we don't know a significantly better way than brute force 19:07:35 ~vixey: I bet you make a simple langauge and compile it into these turing machine state transition things 19:08:37 mark-t: there are lots of constructs that have been shown to be equivalent in power to turing machines 19:08:54 mark-t: one of them is called a while-language, I think, which is somewhat more like ordinary programming languages 19:10:36 ~vixey: what's happening next time? 19:10:59 mark-t: not me 8) Category:Seminar Category:Computer Science